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I proved the inequality below using Wald's identity and some tricky but easy manipulation, but I cannot do it using the suggestion from the source: "Hint: optional sampling!"

Here is the problem:

$X_1,X_2,\ldots$ are i.i.d. with $P(X_1>0)=1$. Let $S_n=\sum_{i=1}^nX_i$ and for $x>0$ define:

$$T_x=min\{n\geq1:S_n\geq x\}$$

Prove:

$$\frac12E[T_x]\leq\frac{x}{E[\min(X_1,x)]}\leq E[T_x]$$


My solution (shown here partially):

Rewrite as: $$x\leq E[\min(X_1,x)]E[T_x]\leq 2x$$

The middle expression is equal to $$E\left[\sum_{i=1}^{T_x}\min(X_i,x)\right]$$ by Wald's identity. Thus we could just prove:

$$x\leq \sum_{i=1}^{T_x}\min(X_i,x)\leq 2x,$$

which is not that hard.


Now how do they do it using optional sampling?

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1 Answer 1

Adaptation of this

Proof of Wald identity using Optional sampling,

may works.

In fact, it's a clever move to know Wald identity.
Otherwise, one needs to derive it using OST, to establish the equality $$ E\left(\sum_{i=1}^{T_x} \min(X_i,x)\right) = E(T_x) E(\min(X_1,x)). $$
Then the solution proceed, as you showed.

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